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40v^2-14v+1=0
a = 40; b = -14; c = +1;
Δ = b2-4ac
Δ = -142-4·40·1
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-6}{2*40}=\frac{8}{80} =1/10 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+6}{2*40}=\frac{20}{80} =1/4 $
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